Hess Law Problem 2

Question

Consider the following reactions:

C(s) + 2H2O(g) --> CO2(g) + 2H2(g) Enthalpy = +163.0 kJ/mole

C(s) + ½O2(g) ---> CO(g) Enthalpy = -121.0 kJ/mole

H2O(g) ---> H2O(l) Enthalpy = -40.7 kJ/mole

CO(g) + ½O2(g) ---> CO2(g) Enthalpy = -283.0 kJ/mole

Calculate the Enthalpy for the reaction H2(g) + ½ O2(g) ---> H2O(l)

Answer 1

I find that the best way to do these problems is to find unique species in the set of reactions that are needed in a given quantity in the target reaction:

The only way to get H₂(g) is to reverse and halve the 1st equation (-1/2 rxn 1)

The only way to get H₂O(l) is to add eqn 3 as is (1 x eqn 3)

The O₂ is included in 2 equations, so instead, let's remove the CO₂ brought in by rxn 1 by cancelling -1/2CO₂ which is the coefficient of CO₂ in reaction 1 or you can think of it as 1/2 mole on the reactant side) is cancelled by adding 1/2 of reaction 3 (+1/2 of rxn 4)

Eliminate the 1/2 CO just introduced by rxn 4 addition on the reactant side by adding 1/2 rxn 2.

You can rewrite the reactions if you wish to check if the 02 is correct. (or 1/2(1/2O2) + 1/2(1/2O2) = 1/2 O2)

You now have the way the reactions combine to form the target reaction. The heats of reaction combine in the same way (Hess's Law)